let $\forall x \neq 0 \ \ \ f(x+\frac{1}{x})=x^2-\frac{1}{x^2}$ then find the $f(x)$
My try :
$$x^2-\frac{1}{x^2}=(x+\frac{1}{x})(x-\frac{1}{x})$$
And $$(x-\frac{1}{x})^2 =(x+\frac{1}{x})^2-4$$
so we have :
$$f(t)=\pm t \sqrt{t^2-4}$$
it is right ?
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Almot1960
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The "so we have" part and following I didn't understand...and your function is defined only for $;|t|\ge2;$ , whereas the given function doesn't have this condition. Yet... – DonAntonio Dec 16 '17 at 21:03
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@DonAntonio Makes sense to me. we have $t = x + 1/x$, so $x-1/x = \pm\sqrt{t^2-4}$ – Dylan Dec 16 '17 at 21:09
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How about $f(t) = t^2-2$? – md2perpe Dec 16 '17 at 21:10
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@md2perpe That would only work if it was $x^2 + 1/x^2$ – Dylan Dec 16 '17 at 21:11
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@Dylan Well, what I was trying the OP to come up with is that the equation $;x+\frac1x=\alpha;$ has a real solution iff $;|\alpha|\ge2;$ ....so the "condition in his function isn't in fact different from the original function's... – DonAntonio Dec 16 '17 at 21:11
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The condition $|x| \ge 2$ isn't really important (or given). All we need is $x + 1/x$ exists for $\forall x \ne 0$ and that $f(x+1/x)$ exists for every value of $x+1/x$ – Dylan Dec 16 '17 at 21:13
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Ah, I see now that I missed the minus sign in the right hand side. – md2perpe Dec 16 '17 at 21:27
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1A similar but different question Given $f(x+1/x) = x^2 +1/x^2$, find $f(x)$ – Jean Marie Dec 16 '17 at 21:53
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Such function cannot exist. Take $x=2$, then we have $$f(2+1/2)= 4-1/4= 15/4$$ Now take $x=1/2$, then we have $$f(2+ 1/2)=1/4-4=-15/4$$But $-15/4\neq 15/4$. Hence such function cannot exist.
Shashi
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